CODEWITHSUNDEEP
scalapattern-matching
Noel Kennedy
Noel Kennedy

Reputation: 12258

Why are both these scala pattern matching extractor calls valid syntax?

tl;dr; It's because of infix operators, nothing to do with extractors

I've been playing around with extractors in scala and found the fact that both the cases below are valid syntax a bit puzzling?! What's the story? 

class Foo(val items:List[String], val number:Int)

object FooEnough {
  def unapply(f:Foo):Option[(List[String], Int)] = if(f.number > 2) Some(f.items, f.number) else None
}

val foo = new Foo("a" :: "b" :: "c" :: Nil, 3)
val matches:String = foo match {
  case FooEnough("a" :: "b" :: "c" :: Nil, 3) => "first case"
  case "a" :: "b" :: "c" :: Nil FooEnough 4 => "second case"
}
//matches == "first case"

If I create another extractor, which extracts 3 parameters, how am I meant to call this using the syntax as in the 2nd case above?

object FooSpectrum {
  def unapply(f:Foo):Option[(String, List[String], Int)] =
    if(f.number > 2) Some("zx", f.items, f.number) else None
}

foo match {
  case FooSpectrum("zx", "a" :: "b" :: "c" :: Nil, 3) => "first case" //compiles
  case "zx" FooSpectrum "a" :: "b" :: "c" :: Nil 4 => "second case" //doesn't compile
}

Upvotes: 2

Views: 279

Answers (1)

drexin
drexin

Reputation: 24423

This is because of the infix operator magic scala provides. You can also write String Map Int instead of Map[String,Int]. This only works for binary types/functions, because it would make no sense for more than 2 args and would also be hard to parse.

Upvotes: 4

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